MCQ
જો $y = {\sin ^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)$, તો ${{dy} \over {dx}} = . . .$
- A${2 \over {1 - {x^2}}}$
- B${1 \over {1 + {x^2}}}$
- ✓$ \pm {2 \over {1 + {x^2}}}$
- D$ - {2 \over {1 + {x^2}}}$
Put $x = \tan \theta $ ==> $\theta = {\tan ^{ - 1}}x$
$\therefore y = {\sin ^{ - 1}}\cos 2\theta = \frac{\pi }{2} \pm 2\theta $
$y = \frac{\pi }{2} \pm 2{\tan ^{ - 1}}x$
$ \Rightarrow $ $\frac{{dy}}{{dx}} = \frac{{ \pm 2}}{{1 + {x^2}}}$.
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